How to use this calculator
- Enter or choose inductance.
- Enter or choose inductance unit.
- Enter or choose total series resistance.
- Enter or choose resistance unit.
- Enter or choose current is.
- The time constant τ updates as you type. Use Copy result or Share to keep it.
Method
τ = L ÷ R; I = V ÷ R; rising: i(t) = I × (1 − e^(−t/τ)); decaying: i(t) = I × e^(−t/τ); f_c = R ÷ (2πL)
- L
- inductance in henries
- R
- total series resistance in ohms
- V
- supply voltage
- I
- steady-state current
- t
- time since the switch
Worked example
- τ = 0.1 H ÷ 10 Ω = 0.01 s = 10 ms.
- Steady-state current I = 12 ÷ 10 = 1.2 A.
- After 10 ms = 1τ: i = 1.2 × (1 − e⁻¹) ≈ 0.7585 A.
A 100 mH coil with 10 Ω in series has τ = 10 ms; 10 ms after switching on 12 V the current is about 0.76 A of its final 1.2 A.
What the result means
Use τ to estimate how quickly a relay, solenoid or motor winding reaches full current, and how long the current takes to die away when switched off (which is why inductive loads need a flyback path).
Assumptions and limits
- Ideal inductor (no core saturation) in series with the resistance entered, switched by a step from a stiff voltage source.
- Decaying current starts from V ÷ R and flows through the same resistance.
- Iron-cored inductors saturate, so L falls at high current and the real rise is faster than calculated.
Questions
Why does the decay need a path?
Interrupting inductor current abruptly produces a large voltage spike (V = L di/dt). A diode or snubber lets the current decay with the time constant instead.
Is the coil resistance included?
Only if you add it. Enter the total of the external resistor and the winding resistance.